How do you use the remainder theorem to find the remainder for each division $(2 x^{3} - 3 x^{2} + x) \div (x - 1)$ $(2x^3-3x^2+x)div(x-1)$ ?

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Answer 1 · 2017-05-17T08:03:06

$0$ $0$

The remainder theorem states that if you divide $P (x)$ $P(x)$ by $(x - a)$ $(x-a)$

the remainder will be $P (a)$ $P(a)$

so dividing $P (x) = (2 x^{3} - 3 x^{2} + x)$ $P(x)=(2x^3-3x^2+x)" "$ by $(x - 1)$ $" "(x-1)$

will give a remainder of $P (1)$ $P(1)$

$P (1) = 2 \times 1^{3} - 3 \times 1^{2} + 1$ $P(1)=2xx1^3-3xx1^2+1$

$P (1) = 2 - 3 + 1 = 0$ $P(1)=2-3+1=0$

which means that in this case $(- 1)$ $(-1)$ is a factor of $P (x)$ $P(x)$

How do you use the remainder theorem to find the remainder for each division (2x^3-3x^2+x)div(x-1)(2x3−3x2+x)÷(x−1)(2x^3-3x^2+x)div(x-1)?